polykin.transport.hmt

Nu_cylinder_free

Nu_cylinder_free(Ra: float, Pr: float) -> float

Calculate the Nusselt number for free convection on a horizontal cylinder.

The average Nusselt number \(\overline{Nu}=\bar{h}D/k\) is estimated by the following expression:

\[ \overline{Nu} = \left(0.6 + \frac{0.387 Ra^{1/6}} {[1 + (0.559/Pr)^{9/16}]^{8/27}}\right)^2 \]

\[ \left[ Ra \lesssim 10^{12} \right] \]

where \(Ra\) is the Rayleigh number and \(Pr\) is the Prandtl number. The properties are to be evaluated at the film temperature.

References

• Churchill, Stuart W., and Humbert HS Chu. "Correlating equations for laminar and turbulent free convection from a horizontal cylinder", International Journal of Heat and Mass Transfer 18, no. 9 (1975): 1049-1053.
• Incropera, Frank P., and David P. De Witt. "Fundamentals of heat and mass transfer", 4th edition, 1996, p. 502.

PARAMETER	DESCRIPTION
Ra	Rayleigh number based on cylinder diameter. TYPE: float
Pr	Prandtl number. TYPE: float

RETURNS	DESCRIPTION
float	Nusselt number.

See also

• Nu_cylinder: related method for forced convection.

Examples:

Estimate the external heat transfer coefficient for a DN25 tube with a surface temperature of 330 K immersed in water with a bulk temperature of 290 K.

>>> from polykin.transport import Nu_cylinder_free
>>> rho = 1.0e3   # kg/m³ (properties at 310 K)
>>> mu = 0.70e-3  # Pa.s
>>> cp = 4.2e3    # J/kg/K
>>> k = 0.63      # W/m/K
>>> beta = 362e-6 # 1/K
>>> D = 33.7e-3   # m (OD from pipe chart)
>>> g = 9.81      # m/s²
>>> Pr = cp*mu/k
>>> Gr = g*beta*(330-290)*D**3/(mu/rho)**2
>>> Ra = Gr*Pr
>>> Nu = Nu_cylinder_free(Ra, Pr)
>>> h = Nu*k/D
>>> print(f"h={h:.1e} W/m².K")
h=1.1e+03 W/m².K

Source code in src/polykin/transport/hmt.py

def Nu_cylinder_free(Ra: float, Pr: float) -> float:
    r"""Calculate the Nusselt number for free convection on a horizontal 
    cylinder.

    The average Nusselt number $\overline{Nu}=\bar{h}D/k$ is estimated by the
    following expression:

    $$ \overline{Nu} = \left(0.6 + \frac{0.387 Ra^{1/6}}
       {[1 + (0.559/Pr)^{9/16}]^{8/27}}\right)^2 $$

    $$ \left[  Ra \lesssim 10^{12} \right] $$

    where $Ra$ is the Rayleigh number and $Pr$ is the Prandtl number. The 
    properties are to be evaluated at the film temperature.

    **References**

    * Churchill, Stuart W., and Humbert HS Chu. "Correlating equations for
      laminar and turbulent free convection from a horizontal cylinder", 
      International Journal of Heat and Mass Transfer 18, no. 9 (1975): 1049-1053.
    * Incropera, Frank P., and David P. De Witt. "Fundamentals of heat and
      mass transfer", 4th edition, 1996, p. 502.

    Parameters
    ----------
    Ra : float
        Rayleigh number based on cylinder diameter.
    Pr : float
        Prandtl number.

    Returns
    -------
    float
        Nusselt number.

    See also
    --------
    - [`Nu_cylinder`](Nu_cylinder.md): related method for forced convection.

    Examples
    --------
    Estimate the external heat transfer coefficient for a DN25 tube with a
    surface temperature of 330 K immersed in water with a bulk temperature of
    290 K.
    >>> from polykin.transport import Nu_cylinder_free
    >>> rho = 1.0e3   # kg/m³ (properties at 310 K)
    >>> mu = 0.70e-3  # Pa.s
    >>> cp = 4.2e3    # J/kg/K
    >>> k = 0.63      # W/m/K
    >>> beta = 362e-6 # 1/K
    >>> D = 33.7e-3   # m (OD from pipe chart)
    >>> g = 9.81      # m/s²
    >>> Pr = cp*mu/k
    >>> Gr = g*beta*(330-290)*D**3/(mu/rho)**2
    >>> Ra = Gr*Pr
    >>> Nu = Nu_cylinder_free(Ra, Pr)
    >>> h = Nu*k/D
    >>> print(f"h={h:.1e} W/m².K")
    h=1.1e+03 W/m².K
    """
    check_range_warn(Ra, 0, 1e12, 'Ra')

    return (0.6 + 0.387*Ra**(1/6) / (1 + (0.559/Pr)**(9/16))**(8/27))**2